Binomial Theorem is a helpful rule in mathematics that lets us expand expressions like (a+b)n, where n is a whole number greater than zero. This theorem tells us that when we expand such expressions, the number of terms we get is always one more than the value of n. For example, if n is 3, we will have 4 terms in the expansion.
Each term in this expansion has a special number called a binomial coefficient, which can be found using a simple arrangement known as Pascal's triangle. In this article, we'll explore Binomial Theorem For Positive Integral Indices in detail.
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Binomial Theorem
Binomial Theorem is a way to expand expressions that are raised to a power, such as (a + b)n, where a and b are any numbers, and n is a non-negative integer (like 2, 3, 4, etc.).
Instead of multiplying (a + b) by itself repeatedly, the binomial theorem gives us a formula to expand it in a quicker way.
Statement of the Binomial Theorem for Positive Integral Indices
The Binomial Theorem states that for any positive integer n the expansion of (a + b)n can be expressed as:
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
where
\binom{n}{k} = \frac{n!}{k! (n-k)!}
In simpler terms, each term in the expansion consists of a coefficient powers of a and powers of b summed over all possible values of the k from the 0 to n.
Formula of Binomial Theorem for Positive Integral Indices
For a general binomial expansion of (a + b)n, the Binomial Theorem gives:
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
To write the general form of the expansion with the first 5–6 terms, we follow the pattern:
(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \binom{n}{3} a^{n-3} b^3 + \binom{n}{4} a^{n-4} b^4 + \binom{n}{5} a^{n-5} b^5 + \cdots
Expanding this we get,
(a + b)^n = a^n + n a^{n-1}b + \frac{n(n-1)}{2!} a^{n-2}b^2 + \frac{n(n-1)(n-2)}{3!} a^{n-3}b^3 + \frac{n(n-1)(n-2)(n-3)}{4!} a^{n-4}b^4 + \frac{n(n-1)(n-2)(n-3)(n-4)}{5!} a^{n-5}b^5 + \cdots
Expanding Binomials with Positive Integral Indices
We can use the above given formula and expand binomials for various values of n. Here we have discussed binomial expansion for first 8 values of n.
- For n = 0:
(a + b)0 = 1
- For n = 1:
(a + b)1 = a + b
- For n = 2:
(a + b)2 = a2 + 2ab + b2
- For n = 3:
(a + b)3 = a3 + 3a2b + 3ab2 + b3
- For n = 4:
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
- For n = 5:
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
- For n = 6:
(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
- For n = 7:
(a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 + b7
Examples with Solutions
Example 1: Expand (x + 3)4 using the Binomial Theorem.
Solution:
Using the formula
(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k where a = x,b = 3 and n = 4,
(x+3)^4 = \binom{4}{0} x^4 \cdot 3^0 + \binom{4}{1} x^3 \cdot 3^1 + \binom{4}{2} x^2 \cdot 3^2 + \binom{4}{3} x^1 \cdot 3^3 + \binom{4}{4} x^0 \cdot 3^4 Calculating each term:
= 1 \cdot x^4 + 4 \cdot x^3 \cdot 3 + 6 \cdot x^2 \cdot 9 + 4 \cdot x \cdot 27 + 1 \cdot 81 = x4 + 12x3 + 54x2 + 108x + 81
Thus, (x + 3)4 = x4 + 12x3 + 54x2 + 108x + 81.
Example 2: Expand (2x-5)3 using the Binomial Theorem.
Solution:
For (a = 2x), (b = -5) and (n = 3):
(2x-5)^3 = \binom{3}{0} (2x)^3 (-5)^0 + \binom{3}{1} (2x)^2 (-5)^1 + \binom{3}{2} (2x)^1 (-5)^2 + \binom{3}{3} (2x)^0 (-5)^3 Calculating each term:
= 1 \cdot (8x^3) - 3 \cdot (4x^2 \cdot 5) + 3 \cdot (2x \cdot 25) - 1 \cdot 125 = 8x3 - 60x2 + 150x - 125
Thus, (2x-5)3 = 8x3 - 60x2 + 150x - 125.
Example 3: Find the expansion of
Solution:
For (a = x),
(b = \frac{1}{2}) and (n = 2):
\left(x + \frac{1}{2}\right)^2 = \binom{2}{0} x^2 \left(\frac{1}{2}\right)^0 + \binom{2}{1} x^1 \left(\frac{1}{2}\right)^1 + \binom{2}{2} x^0 \left(\frac{1}{2}\right)^2 Calculating each term:
= 1 \cdot x^2 + 2 \cdot x \cdot \frac{1}{2} + 1 \cdot \frac{1}{4}
= x^2 + x + \frac{1}{4} Thus,
\left(x + \frac{1}{2}\right)^2 = x^2 + x + \frac{1}{4} .
Example 4: Expand (3 - y)3 using the Binomial Theorem.
Solution:
For (a = 3), (b = -y) and (n = 3):
(3 - y)^3 = \binom{3}{0} 3^3 (-y)^0 + \binom{3}{1} 3^2 (-y)^1 + \binom{3}{2} 3^1 (-y)^2 + \binom{3}{3} 3^0 (-y)^3 Calculating each term:
= 1 \cdot 27 - 3 \cdot 9 \cdot (-y) + 3 \cdot 3 \cdot y^2 - 1 \cdot y^3 = 27 + 27y + 9y2 - y3
Thus, (3 - y)3 = 27 + 27y + 9y2 - y3.
Example 5: Find the expansion of (x + 4)2.
Solution:
For (a = x), (b = 4) and (n = 2):
(x + 4)^2 = \binom{2}{0} x^2 \cdot 4^0 + \binom{2}{1} x^1 \cdot 4^1 + \binom{2}{2} x^0 \cdot 4^2 Calculating each term:
= 1 \cdot x^2 + 2 \cdot x \cdot 4 + 1 \cdot 16 = x2 + 8x + 16
Thus, (x + 4)2 = x2 + 8x + 16.
Practice Questions
Q1: Expand (2a + 3b)3.
Q2: Find the expansion of (x - 2)4.
Q3: What is the expansion of (5 + y)2?
Q4: Expand (x - 1)3.
Q5: Find the expansion of (3x + 2)3.
Q6: Expand
Q7: What is the expansion of (4x - 5)2?
Q8: Find the expansion of (2 - 3x)3.
Q9: Expand (x + 5)3.
Q10: What is the expansion of
Answer Key
- (2a + 3b)3 = 8a3 + 36a2b + 54ab2 + 27b3
- (x − 2)4 = x4 − 8x3 + 24x2 − 32x + 16
- (5 + y)2 = 25 + 10y + y2
- (x − 1)3 = x3 − 3x2 + 3x − 1
- (3x + 2)3 = 27x3 + 54x2 + 36x + 8
\left(a + \frac{1}{3}\right)^4 = a^4 + 4 \cdot a^3 \cdot \frac{1}{3} + 6 \cdot a^2 \cdot \left(\frac{1}{3}\right)^2 + 4 \cdot a \cdot \left(\frac{1}{3}\right)^3 + \left(\frac{1}{3}\right)^4 - (4x − 5)2 = 16x2 − 40x + 25
- (2 − 3x)3 = 8 − 36x + 54x2 − 27x3
- (x + 5)3 = x3 + 15x2 + 75x + 125
\left(2x - 4\right)^2 = 4x^2 - 4 \cdot 2x \cdot 4 + 16
Conclusion
The Binomial Theorem provides a systematic way to expand binomials raised to the any positive integer power. Its applications extend across the many areas of mathematics making it a fundamental concept. By using the binomial coefficients and understanding the relationship between the powers of the a and b we can efficiently solve problems involving the binomial expansions.
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